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Definite integral

Let f:[0, 1]\rightarrow \mathbb{R} be a function such that

    \[f(x^2)+\sqrt {x} f \left( x^2\sqrt {x} \right) = e^x \quad \text{forall} \;\; x \in [0, 1]\]

Evaluate the integral \int_0^1 f(x)\, \mathrm{d}x.


We multiply the given equation by x. Thus,

\begin{aligned} f \left(x^2 \right)+\sqrt {x} f \left(x^2\sqrt {x} \right) &= e^x \Leftrightarrow x f \left(x^2 \right)+x\sqrt{x} f\left(x^2\sqrt {x}\right) =x e^x \\ &\Leftrightarrow \int_{0}^{1} x f \left ( x^2 \right )\, \mathrm{d}x + \int_{0}^{1} x \sqrt{x} f \left ( x^2 \sqrt{x} \right )\, \mathrm{d}x = \int_{0}^{1} xe^x \, \mathrm{d}x \end{aligned}

Let \mathcal{J}_1=\int_{0}^{1} x f \left ( x^2 \right )\, \mathrm{d}x and \mathcal{J}_2= \int_{0}^{1} x \sqrt{x} f \left ( x^2 \sqrt{x} \right )\, \mathrm{d}x.

We now deal with the first integral:

    \begin{align*} \int_{0}^{1}x f\left ( x^2 \right )\, \mathrm{d}x &\overset{u=x^2}{=\! =\!=\!} \frac{1}{2}\int_{0}^{1} f(u)\, \mathrm{d}u \end{align*}

As for the second integral we have:

    \begin{align*}\int_{0}^{1} x\sqrt{x} f\left ( x^2 \sqrt{x} \right )\, \mathrm{d}x &\overset{u=x^2 \sqrt{x}}{=\! =\! =\! =\! =\!} \frac{2}{5} \int_{0}^{1} f(u)\, \mathrm{d}u \end{align*}


    \[\int_{0}^{1} f(x) \, \mathrm{d}x = \frac{10}{9}\]

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